IPMAT Indore 2024 · Quantitative Ability · Set 2: Five constituencies – parties A, B and C

Step 1 — check the totals. Five constituencies with 20 voters each is 100 votes, and 49+35+16=100 ✔ Every voter is accounted for, so in each seat

{a}_{i}+{b}_{i}+{c}_{i}=20

Step 2 — who won S4 and S5? A won only S1, and C won S2 and S3. So S4 and S5 went to B or C.

To win a 20-voter seat outright a party needs more than both others combined would leave it, i.e. strictly more than half of what remains. Concretely, to win seat i a party needs at least 7 votes (if it had 6, the other two would share 14 and one of them would have at least 7).

Now, C won S2 and S3, so {c}_{2}\ge 7 and {c}_{3}\ge 7, using up at least 14 of C’s 16 votes. That leaves C at most 2 votes for S1, S4 and S5 combined — far too few to win anything. Therefore

\text{B won S4 and S5.}

Step 3 — pin down C. Since {c}_{2}+{c}_{3}\le 16 and both are \ge 7, and C must beat B in those seats, the only arrangement that also leaves A and B enough votes elsewhere is

{c}_{2}={c}_{3}=8,{c}_{1}={c}_{4}={c}_{5}=0

Step 4 — B’s votes. {b}_{1}<{b}_{2}<{b}_{3}<{b}_{4}<{b}_{5} are distinct natural numbers summing to 35. B wins S4 and S5 with {c}_{4}={c}_{5}=0, so {b}_{4}>{a}_{4}=20-{b}_{4}, forcing {b}_{4}\ge 11 and likewise {b}_{5}\ge 12. Also {b}_{2}<{c}_{2}=8 and {b}_{3}<{c}_{3}=8.

An exhaustive check of every possibility leaves exactly one distribution:

Verify: each column sums to 20 ✔, each row to its stated total ✔, B’s row is strictly increasing ✔, and every seat has a unique top scorer ✔

Now read the question. We need the seat where B’s tally is below both A’s and C’s.

Only S2 qualifies: 5<7 and 5<8.

Answer: Option 4 — S2

Shortcut. B is last only where C is strong and B is weak. C has votes in just two seats (S2 and S3), so B can possibly be third only there. In S3 B ties A at 6, so it must be S2.

You can reach this without the full grid: B finishes third only where a third party is actually competitive, which narrows the field to two seats before any arithmetic.

Step 1 — check the totals. Five constituencies with 20 voters each is 100 votes, and 49+35+16=100 ✔ Every voter is accounted for, so in each seat

{a}_{i}+{b}_{i}+{c}_{i}=20

Step 2 — who won S4 and S5? A won only S1, and C won S2 and S3. So S4 and S5 went to B or C.

To win a 20-voter seat outright a party needs more than both others combined would leave it, i.e. strictly more than half of what remains. Concretely, to win seat i a party needs at least 7 votes (if it had 6, the other two would share 14 and one of them would have at least 7).

Now, C won S2 and S3, so {c}_{2}\ge 7 and {c}_{3}\ge 7, using up at least 14 of C’s 16 votes. That leaves C at most 2 votes for S1, S4 and S5 combined — far too few to win anything. Therefore

\text{B won S4 and S5.}

Step 3 — pin down C. Since {c}_{2}+{c}_{3}\le 16 and both are \ge 7, and C must beat B in those seats, the only arrangement that also leaves A and B enough votes elsewhere is

{c}_{2}={c}_{3}=8,{c}_{1}={c}_{4}={c}_{5}=0

Step 4 — B’s votes. {b}_{1}<{b}_{2}<{b}_{3}<{b}_{4}<{b}_{5} are distinct natural numbers summing to 35. B wins S4 and S5 with {c}_{4}={c}_{5}=0, so {b}_{4}>{a}_{4}=20-{b}_{4}, forcing {b}_{4}\ge 11 and likewise {b}_{5}\ge 12. Also {b}_{2}<{c}_{2}=8 and {b}_{3}<{c}_{3}=8.

An exhaustive check of every possibility leaves exactly one distribution:

Verify: each column sums to 20 ✔, each row to its stated total ✔, B’s row is strictly increasing ✔, and every seat has a unique top scorer ✔

Read B’s row. {b}_{2}=5.

Why it cannot be anything else. Suppose {b}_{2}=6 (the value the published key gives). The constraint {b}_{1}<{b}_{2}<{b}_{3}<{b}_{4}<{b}_{5} would then force {b}_{3}\ge 7, {b}_{4}\ge 8, {b}_{5}\ge 9.

But C wins S3, so {c}_{3}>{b}_{3}\ge 7, i.e. {c}_{3}\ge 8; and C wins S2, so {c}_{2}>{a}_{2}=20-6-{c}_{2}, giving {c}_{2}\ge 8. Then {c}_{2}+{c}_{3}\ge 16, which exhausts C’s entire total and forces {c}_{1}={c}_{4}={c}_{5}=0 and {c}_{2}={c}_{3}=8. Hence {b}_{3}=7 exactly.

Now B must win S4 and S5 outright against A with no C votes present, so {b}_{4}\ge 11 and {b}_{5}\ge 12. Summing B’s row:

{b}_{1}+6+7+{b}_{4}+{b}_{5}\ge{b}_{1}+6+7+11+12={b}_{1}+36>35

which exceeds B’s total of 35 for any {b}_{1}\ge 0 — a contradiction. So {b}_{2}=6 is impossible.

The same elimination applied to every candidate value leaves {b}_{2}=5 as the only survivor.

Answer: Option 1 — 5

Note on the published key. The widely circulated answer key marks Option 4 (6) here, and 6 is arithmetically impossible once “distinct natural numbers in increasing order” is applied, as shown above.

The key contradicts its own worked solution. The solution table printed alongside it gives B’s row as 1, 5, 6, 11, 12 — identical to the grid above — and its accompanying text states that B “received 1, 5 and 6 votes in S1, S2 and S3”. By the source’s own working, {b}_{2}=5. The printed letter is a transcription error. The verified answer is 5.

Shortcut. Build B’s row from the top down. B must win S4 and S5 against 20-voter fields, so {b}_{4}\ge 11 and {b}_{5}\ge 12 — that is already 23 of B’s 35, leaving just 12 votes for S1, S2 and S3 combined.

With {b}_{1}<{b}_{2}<{b}_{3} summing to 12 and {b}_{3}<8 (C’s 8 must beat it), the only fit is 1+5+6=12. So {b}_{2}=5.

The 23-votes-at-the-top observation is the whole question — it collapses a five-variable problem to a three-variable one in a single line.

Step 1 — check the totals. Five constituencies with 20 voters each is 100 votes, and 49+35+16=100 ✔ Every voter is accounted for, so in each seat

{a}_{i}+{b}_{i}+{c}_{i}=20

Step 2 — who won S4 and S5? A won only S1, and C won S2 and S3. So S4 and S5 went to B or C.

To win a 20-voter seat outright a party needs more than both others combined would leave it, i.e. strictly more than half of what remains. Concretely, to win seat i a party needs at least 7 votes (if it had 6, the other two would share 14 and one of them would have at least 7).

Now, C won S2 and S3, so {c}_{2}\ge 7 and {c}_{3}\ge 7, using up at least 14 of C’s 16 votes. That leaves C at most 2 votes for S1, S4 and S5 combined — far too few to win anything. Therefore

\text{B won S4 and S5.}

Step 3 — pin down C. Since {c}_{2}+{c}_{3}\le 16 and both are \ge 7, and C must beat B in those seats, the only arrangement that also leaves A and B enough votes elsewhere is

{c}_{2}={c}_{3}=8,{c}_{1}={c}_{4}={c}_{5}=0

Step 4 — B’s votes. {b}_{1}<{b}_{2}<{b}_{3}<{b}_{4}<{b}_{5} are distinct natural numbers summing to 35. B wins S4 and S5 with {c}_{4}={c}_{5}=0, so {b}_{4}>{a}_{4}=20-{b}_{4}, forcing {b}_{4}\ge 11 and likewise {b}_{5}\ge 12. Also {b}_{2}<{c}_{2}=8 and {b}_{3}<{c}_{3}=8.

An exhaustive check of every possibility leaves exactly one distribution:

Verify: each column sums to 20 ✔, each row to its stated total ✔, B’s row is strictly increasing ✔, and every seat has a unique top scorer ✔

Step 1 — reduce to a two-party contest. If A and C merge, each seat becomes alliance versus B. Since every seat has exactly 20 voters,

\text{Alliance votes in seat}i={a}_{i}+{c}_{i}=20-{b}_{i}

Step 2 — state the winning condition. The alliance wins seat i when

20-{b}_{i}>{b}_{i}\Leftrightarrow{b}_{i}<10

So the question reduces entirely to: how many of B’s five tallies are below 10?

Step 3 — count.

\text{3 seats}

A stronger statement. Even without pinning down the full grid, B’s row must contain exactly three values below 10: B needs {b}_{4}\ge 11 and {b}_{5}\ge 12 to win its two seats, which leaves at most 35-23=12 votes for the other three seats — so all three of {b}_{1},{b}_{2},{b}_{3} are necessarily below 10. The answer is 3 under every reading of the constraints.

Answer: Option 1 — 3

Note on the published key. The circulated key marks Option 4 (2). For the alliance to win only 2 seats, three of B’s tallies would have to be at least 10, which would require {b}_{3}+{b}_{4}+{b}_{5}\ge 10+11+12=33, leaving under 2 votes for {b}_{1}+{b}_{2} — incompatible with distinct increasing values and with C’s ability to win S2 and S3.

As with Q42, the key contradicts its own worked solution: the table printed in the same document gives B’s row as 1, 5, 6, 11, 12, from which the alliance totals are 19, 15, 14, 9, 8 and it wins S1, S2 and S3 — three seats. The verified answer is 3.

Shortcut. One inequality replaces the whole table. With only two contestants and 20 voters, the alliance wins seat i exactly when {b}_{i}<10. So count how many of B’s numbers are single-digit.

B spends at least 11+12=23 votes winning S4 and S5, leaving at most 12 for the remaining three seats — so all three must be under 10. Answer 3, reachable in two lines without ever constructing A’s or C’s rows.

The general trick: in an alliance question, convert “who wins” into a threshold on the one party left out. It is almost always a shorter route than recomputing every seat.

Step 1 — check the totals. Five constituencies with 20 voters each is 100 votes, and 49+35+16=100 ✔ Every voter is accounted for, so in each seat

{a}_{i}+{b}_{i}+{c}_{i}=20

Step 2 — who won S4 and S5? A won only S1, and C won S2 and S3. So S4 and S5 went to B or C.

To win a 20-voter seat outright a party needs more than both others combined would leave it, i.e. strictly more than half of what remains. Concretely, to win seat i a party needs at least 7 votes (if it had 6, the other two would share 14 and one of them would have at least 7).

Now, C won S2 and S3, so {c}_{2}\ge 7 and {c}_{3}\ge 7, using up at least 14 of C’s 16 votes. That leaves C at most 2 votes for S1, S4 and S5 combined — far too few to win anything. Therefore

\text{B won S4 and S5.}

Step 3 — pin down C. Since {c}_{2}+{c}_{3}\le 16 and both are \ge 7, and C must beat B in those seats, the only arrangement that also leaves A and B enough votes elsewhere is

{c}_{2}={c}_{3}=8,{c}_{1}={c}_{4}={c}_{5}=0

Step 4 — B’s votes. {b}_{1}<{b}_{2}<{b}_{3}<{b}_{4}<{b}_{5} are distinct natural numbers summing to 35. B wins S4 and S5 with {c}_{4}={c}_{5}=0, so {b}_{4}>{a}_{4}=20-{b}_{4}, forcing {b}_{4}\ge 11 and likewise {b}_{5}\ge 12. Also {b}_{2}<{c}_{2}=8 and {b}_{3}<{c}_{3}=8.

An exhaustive check of every possibility leaves exactly one distribution:

Verify: each column sums to 20 ✔, each row to its stated total ✔, B’s row is strictly increasing ✔, and every seat has a unique top scorer ✔

Read A’s row. {a}_{5}=8.

Direct derivation, without the full grid. In S5, C has no votes ({c}_{5}=0, since C’s entire 16 votes are consumed by S2 and S3). So

{a}_{5}=20-{b}_{5}

B’s row is 1,5,6,11,12, so {b}_{5}=12 and

{a}_{5}=20-12=8

Cross-check against A’s total: 19+7+6+9+8=49 ✔

Answer: Option 4 — 8

Shortcut. C is absent from S5, so A and B split all 20 votes. That makes {a}_{5}=20-{b}_{5}, and {b}_{5} is B’s largest value, 12. Hence {a}_{5}=8.

Recognising which seats C contests is the single most useful observation in this set — C’s 16 votes are entirely spent on S2 and S3, which turns S1, S4 and S5 into simple two-party seats where every tally is just 20 minus the other.

Step 1 — check the totals. Five constituencies with 20 voters each is 100 votes, and 49+35+16=100 ✔ Every voter is accounted for, so in each seat

{a}_{i}+{b}_{i}+{c}_{i}=20

Step 2 — who won S4 and S5? A won only S1, and C won S2 and S3. So S4 and S5 went to B or C.

To win a 20-voter seat outright a party needs more than both others combined would leave it, i.e. strictly more than half of what remains. Concretely, to win seat i a party needs at least 7 votes (if it had 6, the other two would share 14 and one of them would have at least 7).

Now, C won S2 and S3, so {c}_{2}\ge 7 and {c}_{3}\ge 7, using up at least 14 of C’s 16 votes. That leaves C at most 2 votes for S1, S4 and S5 combined — far too few to win anything. Therefore

\text{B won S4 and S5.}

Step 3 — pin down C. Since {c}_{2}+{c}_{3}\le 16 and both are \ge 7, and C must beat B in those seats, the only arrangement that also leaves A and B enough votes elsewhere is

{c}_{2}={c}_{3}=8,{c}_{1}={c}_{4}={c}_{5}=0

Step 4 — B’s votes. {b}_{1}<{b}_{2}<{b}_{3}<{b}_{4}<{b}_{5} are distinct natural numbers summing to 35. B wins S4 and S5 with {c}_{4}={c}_{5}=0, so {b}_{4}>{a}_{4}=20-{b}_{4}, forcing {b}_{4}\ge 11 and likewise {b}_{5}\ge 12. Also {b}_{2}<{c}_{2}=8 and {b}_{3}<{c}_{3}=8.

An exhaustive check of every possibility leaves exactly one distribution:

Verify: each column sums to 20 ✔, each row to its stated total ✔, B’s row is strictly increasing ✔, and every seat has a unique top scorer ✔

Read A’s row and take the minimum.

The smallest is 6, in S3.

Answer: Option 3 — S3

Shortcut. A is squeezed hardest where B and C are both strongest. C takes 8 in each of S2 and S3, so A’s worst seats are those two; between them, B polls more in S3 (6) than in S2 (5), leaving A with less. Hence S3.

A quick sanity check: A’s five values must sum to 49 with 19 already spent on S1, so the other four average 7.5 — consistent with the row 7,6,9,8.